A toy is in the shape …

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Sia ? 6 years, 4 months ago
Given the base radius of the cone and the hemisphere are equal.
Diameter of hemisphere = 7 cm
{tex}\Rightarrow{/tex}Radius of hemisphere = 3.5 cm
Radius of the cone = Radius of the hemisphere = 3.5 cm
Let H be the total height of the top.
H = h + r
H = h + 3.5 ...(where h is the height of the cone.) .....(i)
Now,
Volume of toy = Volume of cone + Volume of hemisphere
{tex}\Rightarrow 231 = \left( \frac { 1 } { 3 } \pi r ^ { 2 } h \right) + \left( \frac { 2 } { 3 } \pi r ^ { 3 } \right){/tex}
{tex}\Rightarrow 231 = \frac { 1 } { 3 } \pi r ^ { 2 } ( h + 2 r ){/tex}
{tex}\Rightarrow 231 = \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times 3.5 \times 3.5 ( h + 2 \times 3.5 ){/tex}
{tex}\Rightarrow 231 = \frac { 269.5 } { 21 } ( h + 7 ){/tex}
{tex}\Rightarrow h + 7 = \frac { 4851 } { 269.5 }{/tex}
{tex}\Rightarrow{/tex} h + 7 = 18
{tex}\Rightarrow{/tex} h = 11 cm
{tex}\Rightarrow{/tex} Height of the toy = h + r
= 11 + 3.5 ...(From (i))
= 14.5 cm
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